Modified Euler (Heun's method): Visual walkthrough

4.8.23 · D2 · Maths › Numerical Methods › Modified Euler (Heun's method)

Ek picture-by-picture build of the corrector formula, shuruaat se — "ek slope aakhir curve ke saath karta kya hai?" — har symbol pehle earn hota hai, phir aata hai.

Hum ek hi cheez solve kar rahe hain: ek curve $y (x)$ jis ki kisi bhi point par steepness ek rule se tay hoti hai. Woh rule likhi jaati hai $y^{'} = f (x, y)$ . Pehle isse achhi tarah samjhte hain, baaki sab baad mein.

Step 1 — Ek slope ek direction hai jisme chalna hai

KYA. Starting point $(x_{0}, y_{0})$ par machine hume ek number $k_{1} = f (x_{0}, y_{0})$ deti hai. Woh number ek slope hai: yeh batata hai apne paon kis taraf modhne hain.

KYU. $k_{1}$ ki slope ka matlab hai "har $1$ step daayein ke liye, $k_{1}$ upar jao." Toh agar hum $h$ distance aage badhte hain, altitude $h \cdot k_{1}$ se badlti hai. Yahi woh ek sachchi cheez hai jo hum pehle instant mein jaante hain.

PICTURE. Green curve woh sach jawaab hai jo hum abhi dekh nahin sakte. Shuru mein hum sirf apne paon ke neeche ki slope feel kar sakte hain — yellow arrow. Uss arrow ko extend karna ek seedha andaaza hai.

$k_{1} = slope machine f (start x x_{0}, start y y_{0})$

Har symbol: $f$ pair $(x_{0}, y_{0})$ leta hai aur wahan ki steepness return karta hai. Bas itna hi.

Step 2 — Plain Euler: start-slope pe commit karo (aur drift ho jao)

KYA. Poora step sirf $k_{1}$ use karke chalo: $y_{1}^{(p)} = y_{0} + h k_{1}$

KYU. Yeh sabse sasta possible move hai — hum pehle se $k_{1}$ rakhte hain, toh bas usi ke peeche chalo. Superscript $(p)$ isse ek predictor mark karta hai (rough draft, final answer nahin).

PICTURE. Yellow seedhi line start-slope ko bilkul follow karti hai. Lekin green true curve mod leti hai — iski slope badlti rehti hai. Toh Euler yellow dot par land karta hai, aur true endpoint green dot hai. Vertical red gap error hai.

$y_{1}^{(p)} = start altitude y_{0} + step forward h \cdot start slope k_{1}$

Step 3 — Wahan slope dekhna jahan Euler sochta hai hum land karte hain

KYA. Machine ko predicted endpoint $(x_{1}, y_{1}^{(p)})$ do taaki doosri slope mile: $k_{2} = f (x_{1}, y_{1}^{(p)}) = f (x_{0} + h, y_{0} + h k_{1})$

KYU. Hum jaanna chahte the ki step ke end par trail kitni steep hai — lekin jab tak hum end ki altitude guess nahin karte, machine se end ke baare mein nahin pooch sakte. Step 2 ka draft $y_{1}^{(p)}$ bilkul wahi guess hai. Toh $k_{2}$ exit-slope ka hamara best estimate hai.

PICTURE. Blue arrow yellow (predicted) endpoint par baith ke exit-slope $k_{2}$ dikhata hai. Dhyan do yeh $k_{1}$ se alag taraf point karta hai — curve ke hisaab se steeper ya shallower.

$k_{2} = f (end x x_{0} + h, predicted end y y_{0} + h k_{1})$

Step 4 — Do slopes ko average kyun karo? (trapezoid idea)

KYA. Ab hamare paas do slopes hain: $k_{1}$ start par, $k_{2}$ end par. Heun inke average se chalta hai: $slope used = \frac{k _{1} + k _{2}}{2}$

KYU. Sach par wapas jao. Step par altitude ka change slope-curve ka area hai $x_{0}$ se $x_{1}$ tak: $y (x_{1}) - y (x_{0}) = \int_{x_{0}}^{x_{1}} f (x, y) d x .$ Woh integral sign $\int$ ka matlab hai "step bhar ke saare chhote slope-contributions ko jodo." Hum ise exactly compute nahin kar sakte (andar $y$ nahin jaante), toh hum slope-curve ke neeche ke region ko ek trapezoid se approximate karte hain: ek shape jiska left par true start-height $k_{1}$ hai, right par estimated end-height $k_{2}$ , upar se seedhi line se juda hua.

PICTURE. Baayein: true slope change karti hai (green). Daayein: hum ise ek trapezoid se replace karte hain jiska left edge $k_{1}$ hai, right edge $k_{2}$ . Trapezoid ka area uski width $h$ times uski average height $\frac{k _{1} + k _{2}}{2}$ hai.

$\int_{x_{0}}^{x_{1}} f d x \approx width h \times average height \frac{k _{1} + k _{2}}{2} = \frac{h}{2} (k_{1} + k_{2})$

Yeh bilkul Trapezoidal Rule hai disguise mein. Ek trapezoid ek gently-curving graph ko Euler ke single rectangle se kahin zyada achhi tarah fit karta hai — yehi poora reason hai Heun Euler ko beat karta hai.

Step 5 — Corrector assemble karo

KYA. Trapezoid area ko start altitude mein jodo: $y_{1} = y_{0} + \frac{h}{2} (k_{1} + k_{2})$

KYU. Integral hai hi altitude change; trapezoid hai hi uska estimate; $y_{0}$ mein add karna corrected endpoint par le jaata hai. Ise corrector isliye kehte hain kyunki yeh Step 2 ke crude predictor ko repair karta hai.

PICTURE. Teen seedhi lines overlay ki gayi hain: yellow (Euler, $k_{1}$ follow karta hai), blue (exit-slope $k_{2}$ ), aur red average line jis ki slope $\frac{k _{1} + k _{2}}{2}$ hai. Red endpoint almost exactly green true curve par baitha hai — drift almost gone.

$y_{1} = start y_{0} + half, kyunki do slopes averaged \frac{h}{2} (start slope k_{1} + peeked end slope k_{2})$

Step 6 — Edge case: flat start ( $k_{1} = 0$ )

KYA. Consider karo $y^{'} = - 2 x y$ , $y (0) = 1$ , $h = 0.2$ . Yahan $k_{1} = f (0, 1) = - 2 (0) (1) = 0$ .

KYU yeh matter karta hai. $k_{1} = 0$ hone par predictor hilta nahin: $y^{(p)} = 1 + 0.2 (0) = 1$ . Plain Euler yahan forever ruk jaata aur $y_{1} = 1$ report karta — usse kabhi pata nahin chalta ki curve neeche jhukne wali hai. Yeh degenerate case hai jahan Euler sabse zyada andha hota hai.

PICTURE. Start-arrow (yellow) bilkul horizontal hai. Lekin aage peek karne par, $k_{2} = f (0.2, 1) = - 2 (0.2) (1) = - 0.4$ — ek real downward slope (blue). $0$ aur $- 0.4$ ko average karne se $- 0.2$ milta hai, aur Heun neeche utarta hai. Peek flat-start disaster ko bachata hai.

$y_{1} = 1 + \frac{0.2}{2} (0 + (- 0.4)) = 1 - 0.04 = 0.96$

Exact: $y = e^{- x^{2}} = e^{- 0.04} = 0.9608$ . Heun ka error $0.0008$ ; Euler $1.0$ par stuck tha aur error $0.039$ — karib $50 \times$ bura. Dekho Order of Accuracy and Step Size.

KYA. Corrected $y_{1}$ predictor se better altitude hai. Toh hum re-peek kar sakte hain: $k_{2}$ ko $y_{1}^{(p)}$ ki jagah $y_{1}$ use karke recompute karo, phir dobara correct karo.

KYU. Trapezoidal equation implicit hai — true $y_{1}$ dono sides par aata hai. Ek correction solution ki taraf ek fixed step hai; iterate karna ise aur karib le jaata hai. $y^{'} = x + y$ , $y (0) = 1$ , $h = 0.1$ ke liye: pehla corrector $1.11$ deta hai, re-peek $k_{2} = f (0.1, 1.11) = 1.21$ , dobara correct karo $\to 1.1105$ . Exact answer $1.11034$ hai.

PICTURE. Red endpoint ek baar aur green curve ki taraf thoda nudge karta hai — ek chhoti si final polish. Yeh Heun ko Predictor-Corrector Methods ke wider family se jodhta hai.

Ek-picture summary

Sab ek saath: start slope $k_{1}$ (yellow), peeked end slope $k_{2}$ (blue), averaged red walk green truth par land karti hai, aur trapezoid (Step 4) neeche shaded hai taaki yaad rahe average kahaan se aaya.

Recall Feynman: poori walkthrough apne words mein batao

Tum ek curving trail par hiking kar rahe ho. Step 1: apne paon ke neeche slope feel karo — woh hai $k_{1}$ . Step 2: naively usi direction mein seedha chalo kuch der ke liye; lekin trail curve ho gayi, toh tum drift ho gaye — woh hai plain Euler ka error. Step 3: commit karne ki jagah, ek practice step lo jahan tum land karte, aur wahan slope feel karo — woh hai $k_{2}$ . Step 4: interval par actual climb woh area hai changing-slope graph ke neeche; ek trapezoid ( $k_{1}$ se $k_{2}$ tak flat top) us area ka estimate karta hai, aur ek trapezoid ka area width times uski dono heights ka average hota hai. Step 5: toh start-slope aur peeked end-slope ka average use karke chalo, step se multiply karo — aur tum almost exactly real trail par land karte ho. Step 6: jab bhi start bilkul flat ho aur Euler freeze kar jata, peek ahead aane wali descent pakad leta hai. Step 7: agar aur careful rehna ho, apni improved landing spot se dobara peek karo aur ek baar aur nudge karo.

Active Recall

Heun do slopes average kyun karta hai ek ki jagah?

Kyunki true altitude change slope-curve ke neeche ka area hai, aur ek trapezoid (end heights ka average) us area ko ek rectangle se kahin behtar approximate karta hai.

\frac{h}{2} (k_{1} + k_{2})

formula konsi geometric shape deta hai?

Ek trapezoid — width

h

times average height

\frac{k _{1} + k _{2}}{2}

Flat-start case mein jab

k_{1} = 0

ho, Euler freeze kare toh Heun ko kya bachata hai?

Peeked end-slope

k_{2} = f (x_{1}, y_{0})

phir bhi curve bending detect kar leta hai, toh average nonzero rehta hai.

Connections

Modified Euler (Heun's method) — woh parent jise yeh page unpack karta hai.
Euler's Method — Steps 1–2 exactly plain Euler hain (the predictor).
Trapezoidal Rule — Step 4 yahi rule hai slope integral par apply ki gayi.
Runge-Kutta Methods — assembled formula RK2 hai.
Order of Accuracy and Step Size — kyun trapezoid $O (h^{2})$ deta hai.
Predictor-Corrector Methods — Step 7 corrector ko iterate karta hai.
Taylor Series Methods — second-order match prove karta hai.

Visual walkthrough — Modified Euler (Heun's method)